Question

Energy released in fusion of 1 kg of deuterium nuclei(a) 8 ×10¹³ J(b) 6 ×10²⁷ J(c) 2 × 10⁷ KwH
(d) 8 × 10²³ MeV

Answer 1

To find out the answer of the above mentioned question, first thing to do is to consider the deuterium reaction fusion.

The reaction would be:

1H2 + 1H2 —> 2He3 + 0n1+3.27 MeV

When we apply the energy release formula and substitute values, we will get following answer:

E = 6.02 × 1023 × 103 × 3.27 × 1.6 × 10–13 / 2 × 2

=> E ∼ 9 × 1013 J,

Therefore, we can consdier that ( d ) - 8 × 10²³ MeV is a correct answer.